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The false statement among the following is (1) heating \(1,1,2,2\) -tetrabromoalkane is heated with zinc powder in alcohol gives alkyne (2) 1-alkynes liberate hydrogen with sodium in liquid ammonia (3) Acetylene when treated with \(\mathrm{Cl}_{2}\) in \(\mathrm{CCl}_{4}\) forms lewisite (4) Terminal dihalides on heating with zinc or sodium forms cycloalkanes

Short Answer

Expert verified
The false statement is (3).

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01

- Analyze 1,1,2,2-tetrabromoalkane reaction with zinc powder

When 1,1,2,2-tetrabromoalkane is heated with zinc powder in alcohol, a dehalogenation reaction occurs, leading to the formation of an alkyne. Hence, statement (1) is true.
02

- Analyze 1-alkynes reaction with sodium in liquid ammonia

1-alkynes react with sodium in liquid ammonia to liberate hydrogen gas, confirming that statement (2) is true.
03

- Analyze acetylene reaction with \(\text{Cl}_2\) in \(\text{CCl}_4\)

When acetylene is treated with chlorine (\(\text{Cl}_2\)) in carbon tetrachloride (\(\text{CCl}_4\)), it forms tetrachloroethane, not lewisite. Thus, statement (3) is false.
04

- Analyze terminal dihalides heating with zinc or sodium

When terminal dihalides are heated with zinc or sodium, cycloalkanes are formed. Therefore, statement (4) is true.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dehalogenation Reaction
Dehalogenation is a vital reaction in organic chemistry. It involves the removal of halogens (such as chlorine or bromine) from organic compounds. This reaction is crucial for creating multiple bonds in molecules, like forming an alkyne from a tetrabromoalkane. For example, when 1,1,2,2-tetrabromoalkane is heated with zinc powder in alcohol, a dehalogenation reaction occurs. Zinc acts as a reducing agent, removing halogen atoms and facilitating the formation of an alkyne. This type of reaction is fundamental for synthesizing various organic compounds and illustrates the importance of dehalogenation in organic transformations.
Some key points include:
  • The reaction usually requires a heated environment.
  • Zinc powder is often the reducing agent used.
  • The end product, in this case, is an alkyne.
Alkane and Alkyne Reactions
Alkanes and alkynes are hydrocarbons with distinct properties and reactivities.
Alkanes are saturated hydrocarbons, meaning all carbon atoms are connected by single bonds. They undergo reactions like combustion and substitution. Alkynes, on the other hand, are unsaturated hydrocarbons with one or more carbon-carbon triple bonds. This makes them more reactive than alkanes. When 1-alkynes react with sodium in liquid ammonia, they liberate hydrogen gas.
Here's a simplified breakdown of the reaction:
  • The triple bond of the alkyne makes it nucleophilic.
  • Sodium donates electrons, reducing the alkyne.
  • Hydrogen gas is liberated as a by-product.
Understanding these reactions helps in planning organic syntheses and predicting product outcomes in laboratory settings.
Nucleophilic Addition
Nucleophilic addition reactions are a core aspect of organic chemistry. These reactions involve a nucleophile (a species that donates an electron pair) attacking an electrophile (a species that accepts an electron pair). In the context of the previous exercise, acetylene (a type of alkyne) reacts with chlorine (an electrophile) to form new products. When acetylene is treated with chlorine in carbon tetrachloride, tetrachloroethane is formed, not lewisite.
Key points to remember:
  • Nucleophiles are electron-rich and seek positive centers.
  • Electrophiles are electron-deficient and seek electrons.
  • The result is the formation of new chemical bonds and products.
These reactions are pivotal in forming complex molecules from simpler ones and are widely used in synthetic organic chemistry.
Cycloalkane Formation
Cycloalkane formation is another interesting organic reaction. This involves the creation of ring-structured alkanes. Terminal dihalides (compounds with two halogen atoms at one end) can be converted into cycloalkanes by heating with zinc or sodium.
Here's how it works:
  • Terminal dihalides undergo intramolecular reactions.
  • The halogens are removed by the reducing agent (zinc or sodium).
  • The resulting structure forms a ring, creating a cycloalkane.
This reaction is essential for forming stable ring structures and is utilized in synthesizing many cyclic compounds in organic chemistry. The ability to form rings expands the variety of molecules that chemists can create, demonstrating the versatility and importance of this reaction.

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Most popular questions from this chapter

The addition of water to propyne in the presence of \(\mathrm{HgSO}_{4}-\mathrm{H}_{2} \mathrm{SO}_{4}\) gives (1) \(\mathrm{CII}_{3} \mathrm{CH}=\) CIIOII (2) (3) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (4) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\)

Two organic compounds \(\mathrm{X}\) and \(\mathrm{Y}\) on analysis gave the same percentage composition, namely \(\mathrm{C}=\left(\frac{12}{13}\right) \times\) \(100 \%\) and \(\mathrm{H}=\left(\frac{1}{13}\right) \times 100 \% .\) However, compound "X" decolourises bromine water whilc compound "Y" docs not. Two compounds " \(\mathrm{X}\) " and "Y" may be respectively (1) ethylene and benzene (2) acetylene and ethylene (3) toluene and benzene (4) benzene and styrene

The reduction of an alkyne to alkene using Lindlar's catalyst result in (1) cis addition of hydrogen atoms (2) trans addition of hydrogen atoms (3) a mixture obtained by cis and trans additions of hydrogen which are in equilibrium with each other (4) a mixture obtained by cis and trans additions of hydrogen atoms which are not in equilibrium with each ofher

Which of the following will convert \(\mathrm{HC}=\mathrm{CCH}_{2} \mathrm{CH}_{3}\) to \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3} ?\) (1) \(\mathrm{H}_{2} \mathrm{O} / \mathrm{H}\) (2) \(\mathrm{Hg}^{21} / \mathrm{H}_{2} \mathrm{SO}_{4}\) (3) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (conc) \(/ \mathrm{H}_{3} \mathrm{PO}_{4}\) (4) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{KMnO}_{4}\)

1,1-Dibromopentane on treatment wtih sodamide followed by reaction with \(\mathrm{H}_{3} \mathrm{O}^{+}\) gives (1) 2 -pentyne (2) 3 -mcthyl butyne-1 (3) 1-pentyne (4) \(n\) -propyl cthyne

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